[lucid]

Originally Posted by MDCTFTW

I was thinking about this over lunch ... let me know what you think lucid.

I'm questioning whether the damping coefficient really matters or not when we swap out springs. When we swap in springs they're typically shorter in length and have a higher spring rate. By higher I'm guessing a higher spring rate means it's requires more force for the spring to travel x distance than it would for the stock spring to travel x distance.

Anyway ... say we have 2 identical spring lengths with different spring rates (stock springs A and higher spring rate B). When a spring has a higher spring rate why would that cause more wear than the lower spring rate? Say A & B drive over the same bump at the same speed resulting in the same amount of total force exerted on the system and X amount of travel for A and Y amount of travel for B. Is it safe to say that X will be greater than Y? I would have thought yes.

So if X > Y the dampers travel less for the higher spring rate. How does this cause premature wear on them? I can understand how it could if the springs are shorter and the average travel location on the dampers is different. Maybe I'm missing something.

The transient response of the system is a function of both k (spring rate) and c (damping coeff). Just to clarify, the transient response is how the system would react to an input after it has reached a steady state. So, in the scenario you described, where spring rate kB > is kA, spring A will compress more and than spring B when the springs are installed and the suspension is loaded, and car A will sit lower than car B. That is the steady state. Then you hit the same bump at the same manner with both cars. The transient displacement response of the suspension, how much more the suspension will move, will not only depend on k, but also on c at that point. To understand why that is you need to solve the differential equation I set up above (you substitute -kx for the spring force and -cv for the damper force, and then express acceleration and velocity as time derivates of distance). The nature of the solution (real or complex) depends on a ratio that is termed "the damping ratio", which is:

damping ratio = c / (2 sqrt(k x m))

another parameter is the natural frequency, w0, which is:

w0 = sqrt(k x m)

If the damping ratio is 0, then the system is critically damped, and will return to a steady state the soonest without oscillating. If it is > 0, the system is over-damped, and will take longer to return to steady state (how much longer depends on how overdamped it is), but will still not oscillate. If it is < 0, the system is under-damped, and will overshoot (the spring would compress more than a critically damped system) and oscillate until it returns to steady state at its damped frequency (a function of wo and the damping ratio), but not as fast as a critically damped system.

So, to answer to your question about X and Y, one would need to know not only kA and kB, but also cA and cB. In other words, if car B is under-damped, and car A is critically damped, it is "possible" for Y < X despite kb > kA (car B having stiffer springs).

Now, coming back to why replacing stock springs with aftermarket springs that have a higher k might end up resulting in higher damper forces. If you look at the damping ratio equation above, you will see that if the EDC wants to maintain a specific damping ratio in a given situation--say, it wants critical damping and therefore a damping ratio of 1--and if k is now higher because of the aftermarket springs, then c needs to be higher as well.

Now, we know that Fd = - c x v, and we also know the c of the aftermarket system will be higher. Then, the question is if the velocity of the aftermarket system is higher, equal or lower than the velocity of the stock system. To find that out you can differentiate the displacement response of the critically damped solution, but there might be an easier way. I think you can compare the natural frequencies of the aftermarket and stock systems since that solution is a function of the w0 and time only. So, if the aftermarket system has a higher w0, and if both systems are critically damped, the aftermarket system should experience higher velocities, and therefore, higher damper forces. And since higher k will result in higher w0, that seems to be the case.

I don't do this for a living (not my area of expertise), and am thinking through what I learned in a basics dynamics course 15 years ago. So, you or anyone else is welcome to chime in and critique my thinking on this.

Also, this is a simplified "model" of what really happens. In reality, there are 4 wheels experiencing different inputs but are connected to the same mass, the wheels and suspension parts have mass, stiffness and damping, and inputs vary in size and duration. So, to figure out a real situation, you need to turn to people like Swamp who codes computer simulations, and use those tools. Even then, simulations go so far and sometimes do not explain what needs to be explained.